In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

In two different Young's double-slit experiments,the fringe width is the same when the ratio of wavelengths is $1:2$. If the ratio of the distance between the slits in the two cases is $2:1$,then the ratio of the distance between the slits and the screen in the two experiments is:

In Young's double slit interference experiment,the slit widths are in the ratio $1 : 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is

In Young's double slit experiment, the angular width of a fringe is found to be $0.2^\circ$ on a screen placed $1 \text{ m}$ away. The wavelength of light used is $600 \text{ nm}$. If the entire apparatus is immersed in water of refractive index $4/3$, the angular width of the fringe will be: (in $^\circ$)

In the $Young's$ double slit experiment, the intensity produced by each of the individual slits is $I_0$. The distance between the two slits is $2 \ mm$. The distance of the screen from the slits is $10 \ m$. The wavelength of light is $6000 \ \mathring{A}$. What is the intensity of light on the screen in front of one of the slits?